My roommate here, in the Uni is working on a discrete optimisation problem that is simmilar to siteswap. Since I suck at discrete math, I would like to know the following:

What is the total number of possible states using n balls, and a maximum throw of m? What is an efficient way to calculate them if n = 16 and m = 100 or something like that?

I looked in SS bens book but couldn't find anything. Does anyone know a link?

thanks, Jelmer

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30 Sep 2004 10:32:51

Peter Bone

Re: an actual use for siteswap

Jelmer wrote: > Hi, > > My roommate here, in the Uni is working on a discrete optimisation problem > that is simmilar to siteswap. Since I suck at discrete math, I would like > to know the following: > > What is the total number of possible states using n balls, and a maximum > throw of m? > What is an efficient way to calculate them if n = 16 and m = 100 or > something like that? > > I looked in SS bens book but couldn't find anything. Does anyone know a > link?

I think it's

m nCr n (m combinatorial n nCr = n!/(r!.(n-r)!) )

Because this gives the number of different arrangments of the state sequence - since the state will be m digits long and contain n 1's.

Peter

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30 Sep 2004 11:31:56

juggling jacko

Re: an actual use for siteswap

Just on this subject, i don't understand state/ground state siteswaps. I have read Ben SS guide, but I can't make any sense of it. Could someone please break down states for me.

Thanks heaps!!!

Keep on juggling!!!

<<<<juggling jacko >>>>

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30 Sep 2004 16:25:22

Jack Boyce

Re: an actual use for siteswap

Peter Bone wrote: > Jelmer wrote: > > Hi, > > > > My roommate here, in the Uni is working on a discrete optimisation problem > > that is simmilar to siteswap. Since I suck at discrete math, I would like > > to know the following: > > > > What is the total number of possible states using n balls, and a maximum > > throw of m? > > What is an efficient way to calculate them if n = 16 and m = 100 or > > something like that? > > > > I looked in SS bens book but couldn't find anything. Does anyone know a > > link? > > I think it's > > m nCr n (m combinatorial n nCr = n!/(r!.(n-r)!) ) > > Because this gives the number of different arrangments of the state > sequence - since the state will be m digits long and contain n 1's. > > Peter > > ----== posted via www.jugglingdb.com ==----

That's right. In this case, (100 choose 16) is a big number, so you wouldn't want to try drawing a state diagram!